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• CorrectThe complete SAT Math curriculum, built for your score level?
  Level 6 - Ruby

  650–700

  Custom Regressions0 of 10

  Continue

  60 min

  • Problem D15: Custom Regressions, hard
  • Problem 7b: Custom Regressions, medium, requires Bedrock Pro
  • Problem 7c: Custom Regressions, medium, requires Bedrock Pro
  • Problem 7d: Custom Regressions, hard, requires Bedrock Pro
  • Problem 7f: Custom Regressions, hard, requires Bedrock Pro
  • Problem 8: Radical Models, hard
  • Problem 8b: Radical Models, hard, requires Bedrock Pro
  • Problem 8c: Radical Models, hard, requires Bedrock Pro
  • Problem 44: Points on Exponential Function, medium
  • Problem 44e: Points on Exponential Function, medium, requires Bedrock Pro

  More practice (5)

  • Problem 7: Custom Regressions, hard, free video solution
  • Problem 7e: Custom Regressions, hard, requires Bedrock Pro
  • Problem 8d: Radical Models, hard, requires Bedrock Pro
  • Problem 44h: Points on Exponential Function, hard, requires Bedrock Pro
  • Problem 44i: Points on Exponential Function, hard, requires Bedrock Pro

  Percentages 30 of 5

  30 min

  • Problem D22: Percentages, hard
  • Problem 37: Hard Percentages, hard
  • Problem 37b: Hard Percentages, hard, requires Bedrock Pro
  • Problem 37c: Hard Percentages, hard, requires Bedrock Pro
  • Problem 37h: Hard Percentages, hard, requires Bedrock Pro

  More practice (4)

  • Problem 37d: Hard Percentages, hard, requires Bedrock Pro
  • Problem 37e: Hard Percentages, hard, requires Bedrock Pro
  • Problem 37f: Hard Percentages, hard, requires Bedrock Pro
  • Problem 37g: Hard Percentages, hard, requires Bedrock Pro

  Vertex Form & Quadratic Symmetry0 of 10

  65 min

  • Problem 62b: Vertex Form, medium, requires Bedrock Pro
  • Problem 62d: Vertex Form, medium, requires Bedrock Pro
  • Problem 60c: Quadratic Applications, medium, requires Bedrock Pro
  • Problem 60d: Quadratic Applications, medium, requires Bedrock Pro
  • Problem 60h: Quadratic Applications, hard, requires Bedrock Pro
  • Problem 60: Quadratic Applications, hard
  • Problem 62: Vertex Form, hard
  • Problem 62e: Vertex Form, hard, requires Bedrock Pro
  • Problem 63c: Lowest Possible Sum, hard, requires Bedrock Pro
  • Problem 63: Lowest Possible Sum, hard

  More practice (9)

  • Problem 60b: Quadratic Applications, medium, requires Bedrock Pro
  • Problem 60e: Quadratic Applications, medium, requires Bedrock Pro
  • Problem 60f: Quadratic Applications, hard, requires Bedrock Pro
  • Problem 60g: Quadratic Applications, hard, requires Bedrock Pro
  • Problem 62c: Vertex Form, medium, requires Bedrock Pro
  • Problem 62f: Vertex Form, hard, requires Bedrock Pro
  • Problem 62g: Vertex Form, hard, requires Bedrock Pro
  • Problem 63b: Lowest Possible Sum, hard, requires Bedrock Pro
  • Problem 63d: Lowest Possible Sum, hard, requires Bedrock Pro

  Tangent to a Circle0 of 6

  45 min

  • Problem 85b: Line Tangent to Circle, medium, requires Bedrock Pro
  • Problem 85c: Line Tangent to Circle, medium, requires Bedrock Pro
  • Problem 85: Line Tangent to Circle, medium
  • Problem 86b: Quadrilateral from Tangent Lines, hard, requires Bedrock Pro
  • Problem 86c: Quadrilateral from Tangent Lines, hard, requires Bedrock Pro
  • Problem 86: Quadrilateral from Tangent Lines, hard, free video solution

  More practice (4)

  • Problem 85d: Line Tangent to Circle, medium, requires Bedrock Pro
  • Problem 85e: Line Tangent to Circle, hard, requires Bedrock Pro
  • Problem 86d: Quadrilateral from Tangent Lines, hard, requires Bedrock Pro
  • Problem 87h: Standard Form of a Circle, hard, requires Bedrock Pro

  Margin of Error0 of 4

  15 min

  • Problem 125b: Margin of Error, easy, requires Bedrock Pro
  • Problem 125c: Margin of Error, medium, requires Bedrock Pro
  • Problem 125: Margin of Error, medium
  • Problem 125e: Margin of Error, medium, requires Bedrock Pro

  More practice (1)

  • Problem 125d: Margin of Error, medium, requires Bedrock Pro
• CorrectVideo lessons for 127 core problems?
  
• CorrectConcept guides and drills to build fundamentals?
  The Inscribed Angle Theorem

  3 min

  Two angles that face the same arc

  Here's a circle with center $O$ and three points $P$, $Q$, and $R$ on it.

  

  The $\spotblue{\text{blue}}$ angle has its vertex at the center. That makes it a central angle. A central angle and the arc it cuts off are the same measure: a $70^\circ$ central angle cuts off a $70^\circ$ arc. Think of the angle and the $\spotorange{\text{arc}}$ as one measurement, just drawn at different distances from the center.

  The $\spotred{\text{red}}$ angle has its vertex on the circle. That makes it an inscribed angle. Its two sides are chords, and it opens onto the same orange arc the central angle does.

  The inscribed angle is half

  Here's the same figure with the numbers on.

  

  The $\spotred{35^\circ}$ at $R$ is half of the $\spotblue{70^\circ}$ at $O$. That is always true, for any inscribed angle:

  $$\spotred{\text{inscribed angle}} = \frac{1}{2} \cdot \spotblue{\text{central angle}}$$ $$\spotblue{\text{central angle}} = 2 \cdot \spotred{\text{inscribed angle}}$$

  This is the inscribed angle theorem. It doesn't matter where $R$ sits on the circle. Slide it anywhere along the far side and the red angle stays $\spotred{35^\circ}$, because it still faces the same $70^\circ$ arc.

  You should memorize this for the SAT. It is not on the formula sheet.

  When the arc is half the circle

  Suppose $\overline{PQ}$ passes through the center, so it's a $\spotblue{\text{diameter}}$.

  

  The central angle from $P$ to $Q$ is a straight line, so it's $180^\circ$. Half of that is $90^\circ$, so the inscribed angle at $R$ is a $\spotred{\text{right angle}}$.

  It runs the other way too. If a triangle sits inside a circle with one side a diameter, the angle across from that side is $90^\circ$. Recognize this figure on sight: it's a right triangle, and everything you know about right triangles applies.

• Correct970+ problems with written explanations?
  Angles & Trianglesmedium

  Problem 77k

  Similarity

  

  In the figure shown, $\overline{JM}$ and $\overline{KL}$ intersect at point $P$, $JP = 45$, $LP = 18$, $JK = 40$, and $KP = 60$. What is the length of $\overline{LM}$?

  Note: Figure not drawn to scale.

  • A

    $12$

  • B, correct

    $16$

  • C

    $24$

  • D

    $100$

  Solution

  Scale the triangle downSet up a proportion

  Solution approachScale the triangle downSet up a proportion

  Step 1. The figure marks angle $J$ and angle $L$ with the same $a^\circ$. $\spotgreen{\angle JPK}$ and $\spotgreen{\angle LPM}$ are vertical angles, so they're equal too. Two matching angles are enough for similarity, so triangle $JKP$ is similar to triangle $LMP$, with $J$ corresponding to $L$, $K$ to $M$, and $P$ to $P$.

  

  Step 2. Similar triangles are scaled copies, so every side is multiplied by the same number. Pair the sides by their endpoints. $J$ and $P$ match $L$ and $P$, so $\spotred{\overline{JP}}$ corresponds to $\spotred{\overline{LP}}$. $J$ and $K$ match $L$ and $M$, so $\spotblue{\overline{JK}}$ corresponds to $\spotblue{\overline{LM}}$.

  

  Step 3. The problem gives us $\spotred{JP} = 45$ and $\spotred{LP} = 18$, so triangle $LMP$ is triangle $JKP$ scaled by $\dfrac{18}{45}$. Call that number $k$.

  Step 4. Every side scales by that same $k$, and the problem gives us $\spotblue{JK} = 40$:

  $$\spotblue{LM} = 40k$$

  1$\displaystyle k = \frac{18}{45}$

  2$\displaystyle 40k$

  Graph it hereOpen in Desmos

  $\overline{LM}$ has a length of $16$. Choice B.

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  Practice Test 1

  710

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